Answer
(a) $I_L = 5.0~mA$
(b) $I_L = 20~mA$
Work Step by Step
We can write an expression for the peak current:
$I_L = \frac{\epsilon_0}{\omega ~L}$
$I_L = \frac{\epsilon_0}{2\pi ~f ~L}$
(a) If the emf frequency is doubled, then the peak current decreases by a factor of $\frac{1}{2}$
$I_L = 5.0~mA$
(b) If the emf peak voltage is doubled, then the peak current increases by a factor of $2$
$I_L = 20~mA$
